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So I can expand the numerator by geometric mean. Letting C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n. Let the numerator be called a_{n} and the denominator be b_{n} Is there a way to use this statement so that I could force the original sequence into the form of 1/\left(1+\frac{1}{n}\right)^n
Answer
I would like to use the following lemma:
If \lim_{n\to\infty}a_n=a and a_n>0 for all n, then we have
\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}=a \tag{1}
Let a_n=(1+\frac{1}{n})^n, then a_n>0 for all n and \lim_{n\to\infty}a_n=e. Applying (*) we have
\begin{align} e&=\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}\\ &=\lim_{n\to\infty}\sqrt[n]{\left(\frac{2}{1}\right)^1\left(\frac{3}{2}\right)^2\cdots\left(\frac{n+1}{n}\right)^n}\\ &=\lim_{n\to\infty}\sqrt[n]{\frac{(n+1)^n}{n!}}\\&= \lim_{n\to\infty}\frac{n+1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}+\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} \end{align}\tag{2}
where we use (1) in the last equality to show that
\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0.
It follows from (2) that
\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=\frac{1}{e}.
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